Discussion on Counting Valleys Challenge

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4 days ago+ 0 comments

can you tell me how i can attach this java script steps to describing the path of my travel geek. Thank you in advance for your help!

1 week ago+ 1 comment

JavaScript solution with early exit condition:

function countingValleys(steps, path) {
    // Write your code here
    var i = 0;
    var elevation = 0;
    var valleys = 0;
    
    while(i < steps && steps - i > Math.abs(elevation)){
        if(path[i] == "U")
            elevation++;
        else {
            elevation--;
            if(elevation == -1)
                valleys++;
        }
            
        i++;
    }
    
    return valleys;
}

4 days ago+ 0 comments

I want to use this amazing JavaScript solution on website that this hosted on Wordpress. It will be very hepful for my website stardew valley ios

1 week ago+ 0 comments

public static int countingValleys(int steps, String path) {
int seaLevel = 0, valleys = 0, currentLevel = 0;

for (char step : path.toCharArray()) {
    if (step == 'U') {
        currentLevel++;
    } else {
        currentLevel--;
    }

    // If we just came up to sea level from below, we've exited a valley
    if (currentLevel == 0 && step == 'U') {
        valleys++;
    }
}

return valleys;
}

2 weeks ago+ 0 comments

def countingValleys(steps, path):
    level = 0
    valley = False
    path_values = list(map(direction_value, path))
    valleys = 0
    
    for s in path_values:
        level += s
        if level < 0:
            valley = True
        if valley and level >= 0:
            valleys += 1
            valley = False
            
    return valleys
    
def direction_value(d):
    if d == 'U':
        return 1
    return -1

2 weeks ago+ 0 comments

Here is my c++ solution you can find the video here : https://youtu.be/fgJ-i8RJ1Qw

int countingValleys(int steps, string path) {
    int res = 0, level = 0;
    for(char c : path){
        if(c == 'U'){
            level++;
            if(level == 0) res ++;
        }
        else level--;
    }
    return res;
}

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